A vessel having certain amount of oxygen at certain pressure and temperature develops a very small hole and 4 g of oxygen effuses out. How much hydrogen would have effused out of the same vessel had it been taken at the same pressure and temperature.

Options

A. 1g
B. 16g
C. 0.25g
D. 64g

Answer

To solve the problem of how much hydrogen would have effused out of the same vessel, we can use Graham's law of effusion. Here’s a step-by-step solution:

### Step 1: Understand Graham's Law of Effusion
Graham's law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. Mathematically, it can be expressed as:
\[ \frac{R_1}{R_2} = \sqrt{\frac{M_2}{M_1}} \]
where \( R_1 \) and \( R_2 \) are the rates of effusion of gas 1 (oxygen) and gas 2 (hydrogen), and \( M_1 \) and \( M_2 \) are their respective molar masses.

### Step 2: Identify the Given Information
- Mass of oxygen effused (\( W_1 \)) = 4 g
- Molar mass of oxygen (\( M_1 \)) = 32 g/mol
- Molar mass of hydrogen (\( M_2 \)) = 2 g/mol

### Step 3: Set Up the Equation Using Graham's Law
Since the vessel is at the same pressure and temperature, we can set the rates of effusion in terms of the number of moles:
\[ \frac{n_1}{n_2} = \sqrt{\frac{M_2}{M_1}} \]
Where \( n_1 \) and \( n_2 \) are the number of moles of oxygen and hydrogen, respectively.

### Step 4: Express Moles in Terms of Mass
The number of moles can be expressed as:
\[ n = \frac{W}{M} \]
Thus, we can rewrite the equation:
\[ \frac{\frac{W_1}{M_1}}{\frac{W_2}{M_2}} = \sqrt{\frac{M_2}{M_1}} \]
This simplifies to:
\[ \frac{W_1 \cdot M_2}{W_2 \cdot M_1} = \sqrt{\frac{M_2}{M_1}} \]

### Step 5: Substitute the Known Values
Substituting the known values into the equation:
\[ \frac{4 \cdot 2}{W_2 \cdot 32} = \sqrt{\frac{2}{32}} \]

### Step 6: Simplify the Equation
First, simplify the right side:
\[ \sqrt{\frac{2}{32}} = \sqrt{\frac{1}{16}} = \frac{1}{4} \]
Now, substituting this back into the equation:
\[ \frac{8}{W_2 \cdot 32} = \frac{1}{4} \]

### Step 7: Cross Multiply to Solve for \( W_2 \)
Cross multiplying gives:
\[ 8 \cdot 4 = W_2 \cdot 32 \]
\[ 32 = W_2 \cdot 32 \]
Dividing both sides by 32:
\[ W_2 = 1 \text{ g} \]

### Conclusion
Thus, the mass of hydrogen that would have effused out of the same vessel is **1 gram**.

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